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2 years ago

WIL GIVE BRAINLISEST PLEASE HELP SPEED ACCELERATION GRAPHS LABEL THE GRAPHS PLEASEEeeee

Physics

1 answer:

Varvara68 [4.7K]2 years ago

4 0

Answer:

Take the numbers and fill them in graph.

Plot time as X and speed as Y and afterward join then like a line graph.

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An object’s average density rho is defined as the ratio of its mass to its volume: rho=M/V. The earth’s mass is 5.94×1024kg, and

ladessa [460]

Answer:

The answer to your question is 5.5 x 10¹² kg/km³

Explanation:

Data

density = rho = X

mass = 5.94 x 10²⁴ kg

volume = 1.08 x 10¹² km³

Process

To solve this problem use the density formula and simplify it to find the result.

Formula

density = $\frac{mass}{volume}$

Substitution

density = $\frac{5.94 x 10^{24} }{1.08 x 10^{12}}$

Simplification and result

density = 5.5 x 10¹² kg/km³

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2 years ago

How can two tectonic plates move relative to each other

yarga [219]

They are doing the transform type of tectonic plate movement. This is a process where the plates move sideways. But they move at a rate of a few inches a year. Did that help you?

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3 years ago

A goose waddles 18.0 m north and then 5.00 m east. He then walks 2.00 m south and 2.00 m west. What is the magnitude and directi

olya-2409 [2.1K]

Your answer should be 16.3 m 79.4º east of north

7 0

3 years ago

An ideal spring hangs from the ceiling. A 2.15 kg mass is hung from the spring, stretching the spring a distance d = 0.0895 m fr

Igoryamba

Answer:

The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

Explanation:

Given that,

Mass = 2.15 kg

Distance = 0.0895 m

Amplitude = 0.0235 m

We need to calculate the spring constant

Using newton's second law

$F= mg$

Where, f = restoring force

$kx=mg$

$k=\dfrac{mg}{x}$

Put the value into the formula

$k=\dfrac{2.15\times9.8}{0.0895}$

$k=235.41\ N/m$

We need to calculate the kinetic energy of the mass

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Here, $v = A\omega$

$K.E=\dfrac{1}{2}m\times(A\omega)^2$

Here, $\omega=\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}m\times A^2\sqrt{\dfrac{k}{m}}^2$

$K.E=\dfrac{1}{2}kA^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times235.41\times(0.0235)^2$

$K.E=0.06500\ J$

Hence, The kinetic energy of the mass at the instant it passes back through the equilibrium position is 0.06500 J.

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3 years ago

State the Pascal's principle <br>

krok68 [10]

Answer:

Pascal's principle, also called Pascal's law, in fluid (gas or liquid) mechanics, statement that, in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container

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2 years ago